True or False:
Electrons in an s-orbital are more effective than those in other orbitals at shielding other electrons from the nuclear charge b/c an electron in an s-orbital can penetrate to the nucleus of the atom.
What does "penetrate to the nucleus of the atom" mean? Does this refer to the penetration effect, aka the amount of "pull" electrons feel from the nucleus?
Here's where I found info about the penetration effect:
https://www.quora.com/What-do-we-unders ... f-orbitals
Textbook Question 2.37 part b [ENDORSED]
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Re: Textbook Question 2.37 part b [ENDORSED]
The statement is true because the s-orbital is situated more closely to the nucleus of the atom than p, d, and f orbitals. As a result, the inner electrons that are situated in the s-orbital shield outer electrons from the electrostatic attraction of the positively charged nucleus. In regards to your question about "penetrate to the nucleus of the atom", what that statement is implying is the ability of an electron to get close to the nucleus "feel the pull" that is based on where the electron is located in terms of orbitals; an electron located in the s-orbital can "penetrate to the nucleus", but if an electron is located in the p ,d, and f orbitals it won't "penetrate to the nucleus" due to the shielding effect.
I hope this helps!
I hope this helps!
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Re: Textbook Question 2.37 part b
I also have a question on 2.37 but it's for part d. So an electron in an s orbital has a lower energy than an electron in a p orbital, but it has a higher effective nuclear charge. What exactly is effective nuclear charge and does it have an inverse relationship to the energy of the electron?
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Re: Textbook Question 2.37 part b
Wait I just answered my own question, but in case other people are also confused by this, maybe it will help. Effective nuclear charge has to do with the pull from the nucleus on electrons in their respective orbitals, so an electron in an s orbital will have a stronger attraction to the nucleus than an electron in a p orbital because of shielding properties. This relates to energy because of the equation at the top of page 43, which indicates a decrease in potential energy, the closer the electron is to the nucleus (because of the negative sign in the equation). Therefore, an s orbital electron will have a higher effective nuclear charge and a lower potential energy.
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